Question: The lifespans of seals in a particular zoo are normally distributed. The average seal lives $15$ years; the standard deviation is $3.3$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a seal living longer than $8.4$ years.
Solution: $15$ $11.7$ $18.3$ $8.4$ $21.6$ $5.1$ $24.9$ $95\%$ $2.5\%$ $2.5\%$ We know the lifespans are normally distributed with an average lifespan of $15$ years. We know the standard deviation is $3.3$ years, so one standard deviation below the mean is $11.7$ years and one standard deviation above the mean is $18.3$ years. Two standard deviations below the mean is $8.4$ years and two standard deviations above the mean is $21.6$ years. Three standard deviations below the mean is $5.1$ years and three standard deviations above the mean is $24.9$ years. We are interested in the probability of a seal living longer than $8.4$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $95\%$ of the seals will have lifespans within 2 standard deviations of the average lifespan. The remaining $5\%$ of the seals will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({2.5\%})$ will live less than $8.4$ years and the other half $({2.5\%})$ will live longer than $21.6$ years. The probability of a particular seal living longer than $8.4$ years is ${95\%} + {2.5\%}$, or $97.5\%$.